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The separation of variables is a mathematical technique often used to solve simple differential equations. It involves rearranging a differential equation such that each variable appears on only one side of the equation.
For example, in the given problem, the differential equation is initially presented as:

• \( \frac{dR}{dy} + R = 1 \)

To apply separation of variables, we work to isolate the differential term. By moving all \( R \)-related terms to one side, we achieve this:

• \( \frac{dR}{1-R} = dy \)

Here, the variable \( R \) is completely separated on the left, and the variable \( y \) is isolated on the right. This clear separation allows for the integration of both sides, leading us to the next step in solving the equation. It's a foundational step in transforming a differential equation into a form that is ready to be solved.

Initial conditions are supplementary data for differential equations. They allow us to determine specific solutions rather than general solutions.
Initial conditions usually provide information on the value of the function or its derivative at a particular point. In this exercise, the initial condition \( R(1) = 0.1 \) is provided.
Using the general solution after integrating,

• \( R = 1 - Ce^{-y} \)

inserting the given initial condition helps us solve for the integration constant \( C \).
Substituting \( y = 1 \) and \( R = 0.1 \):

• \( 0.1 = 1 - Ce^{-1} \)

Solving this equation yields \( C = 0.9e \), which, when substituted back, allows us to find the particular solution to this problem. Initial conditions are crucial because they help personalize our differential equation to a specific scenario.

Integration is the mathematical process used to solve a differential equation once the variables have been separated. It essentially involves finding the antiderivative of both sides of the equation.
When we separated the equation \( \frac{dR}{1-R} = dy \), we ended up having to integrate both sides:

• The left side: \( \int \frac{1}{1-R} \, dR \)
• The right side: \( \int \, dy \)

The result of the integration on the left is \(-\ln|1-R|\) and on the right is simply \( y + C \).
This integration leads us to the equation:\[-\ln|1-R| = y + C \] The integration constant \( C \) is important and is calculated using initial conditions. Integration transforms our differential equation into a form where we can precisely solve for the function \( R \) using algebraic techniques.

Exponential functions are critical in solving differential equations, particularly when transforming logarithmic solutions back to solve for the original function. In the result of our integration, we have the expression \(-\ln|1-R| = y + C \).
To solve for \( R \), we need to express this in terms of an exponential equation:

• Transform \(-\ln|1-R|\) back to \( 1-R \) using exponents: \( 1 - R = Ce^{-y} \)

The general solution now indicates that \( R \) changes exponentially with respect to \( y \).
When we applied the initial condition and solved for \( C \), our particular solution manifested as an exponential function:

• \( R = 1 - 0.9e^{1-y} \)

Exponential functions are common in differential equations due to their growth and decay properties, which naturally model many real-world phenomena such as population dynamics and radioactive decay.